What is the integral from 1 to x of tsqrt(y+3) ?

The integral from 1 to x of tsqrt(y+3) is ((x^2-1)sqrt(3+y))/2

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integral from 1 to x of tsqrt(y+3)

\int_{1}^{x} t y + 3 d t

\frac{( x ^{2} - 1 ) 3 + y}{2}

Solution steps

\int_{1}^{x} t y + 3 d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 + y \cdot \int_{1}^{x} t d t

Apply the Power Rule

: [\frac{t ^{2}}{2}]_{1}^{x}

= 3 + y [\frac{t ^{2}}{2}]_{1}^{x}

Compute the boundaries:

\frac{x ^{2}}{2} - \frac{1}{2}

= 3 + y (\frac{x ^{2}}{2} - \frac{1}{2})

Simplify

= \frac{( x ^{2} - 1 ) 3 + y}{2}

\int_{1}^{2} x^{5} ln (x) d x

2 x y d x + (2 + x^{2}) d y = 0

y^{^{''}} - 2 y^{^{'}} = x + 2 e^{x}

\int (x^{2} - \frac{3}{x} + cos (x) - \frac{2}{x}) d x

y = x^{2} - 7

What is the integral from 1 to x of tsqrt(y+3) ?
The integral from 1 to x of tsqrt(y+3) is ((x^2-1)sqrt(3+y))/2